Abstract
We introduce a sound non-wellfounded proof system whose regular (or ‘cyclic’) proofs are complete for (in)equations between regular expressions. We achieve regularity by using hypersequents rather than usual sequents, with more structure in the succedent, and relying on the discreteness of rational languages to drive proof search. By inspection of the proof search space we extract a PSpace bound for the system, which is optimal for deciding such (in)equations.
An extended version of this abstract is available on HAL [12]. This work was supported by the European Research Council (ERC) under the Horizon 2020 programme (CoVeCe, grant agreement No. 678157) and the LABEX MILYON (ANR-10-LABX-0070) of Université de Lyon, within the program “Investissements d’Avenir” (ANR-11-IDEX-0007).
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Notes
- 1.
Here we write \(x \le y\) as a shorthand for \(x+y = y\).
- 2.
This logic is non-commutative because there is no exchange rule, and intuitionistic since there is exactly one formula on the right-hand side.
- 3.
Note that atomicity of e really is required for this, even in the usual rational language model. For instance, we have \({\mathcal L}(a^* a b )\subseteq {\mathcal L}(a^* b )\), but \({\mathcal L}(ab) \nsubseteq {\mathcal L}(b)\).
- 4.
Strictly speaking, we should bracket \(e^n\) as \(e(e(\cdots (ee)))\) and set \(e^0\) to 1.
- 5.
Here we construe multisets as mappings from elements to their multiplicity.
- 6.
A priori, this could still be exponentially many in the size of the end-sequent.
- 7.
Notice that right logical rules do not branch.
- 8.
This argument is akin to applying a cut, which is sound since we are only applying it once, and at the meta-level.
- 9.
Here we mean in the sense that it is identical to a descendant, as in Lemma 20.
- 10.
Notice that the \(*\) rules here correspond in fact to an alternative fixed point definition of \(e^*\): \(\mu x . (1 + e + xx)\).
- 11.
Notice also that while it would be natural to enrich the antecedent structure for \(\cap \) as we did in succedents for \(+\), there is a difficult asymmetry in that \(x(y+z) = xy + xz\) but \(x(y\cap z) \lneq xy \cap xz\).
- 12.
Note that the broader problem of whether cyclic proofs can be simulated by ‘inductive’ proofs for a certain framework has no known general solution, cf. [6].
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Das, A., Pous, D. (2017). A Cut-Free Cyclic Proof System for Kleene Algebra. In: Schmidt, R., Nalon, C. (eds) Automated Reasoning with Analytic Tableaux and Related Methods. TABLEAUX 2017. Lecture Notes in Computer Science(), vol 10501. Springer, Cham. https://doi.org/10.1007/978-3-319-66902-1_16
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